hdu 1026 Ignatius and the Princess I

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1026

题目大意：求用时最短路线并打印路径，n表示用n秒打败怪兽。

题目分析：bfs+优先队列求出最短用时，path数组记录路径，栈输出路径。

代码参考：

#include <queue>#include <stack>#include <cstdio>#include <vector>#include <cstring>#include <algorithm>using namespace std;const int N = 109;priority_queue<int> pq;//优先队列struct Node{    int x, y, step;    friend bool operator < (const Node &a, const Node &b) { //定义排序规则        return a.step > b.step;    }}p[N][N],path[N][N];char edge[N][N];//保存地图bool vis[N][N];int dir[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};int n, m, minStep;int bfs(Node s){    priority_queue<Node> Q;    Q.push(s);    vis[0][0] = 1;    while(!Q.empty()) {        Node head = Q.top();        Q.pop();        if(head.x == n - 1 && head.y == m - 1) { //如果走到了终点，返回最短用时            return head.step;        }        for(int i = 0; i < 4; ++ i) { //枚举4个方向            int x = head.x + dir[i][0];            int y = head.y + dir[i][1];            if(x >= 0 && x < n && y >= 0 && y < m && edge[x][y] != 'X' && !vis[x][y]) { //判定是否可走                vis[x][y] = 1;                path[x][y].x = head.x; //保存路径                path[x][y].y = head.y;                path[x][y].step = head.step + 1;                Node t; //当前点                t.x = x;                t.y = y;                t.step = head.step + 1;                if(edge[x][y] == '.') {                    Q.push(t);                } else {                    t.step = t.step + edge[x][y] - '0'; //加上打怪兽的时间                    Q.push(t);                }            }        }    }    return -1; //如果没有走到终点}void print(){    if(minStep == -1) { //如果没有走到终点        puts("God please help our poor hero.");    } else { //输出最短用时        printf("It takes %d seconds to reach the target position, let me show you the way.\n", minStep);        stack<Node> st; //栈输出        Node t = path[n - 1][m - 1];        Node rmin;        rmin.x = n - 1;        rmin.y = m - 1;        rmin.step = minStep;        st.push(rmin);        while(t.x || t.y) { //如果到了终点            st.push(t); //将路径压入栈内            t = path[t.x][t.y];        }        int k = 1;        while(!st.empty()) {            t = st.top();            st.pop();            printf("%ds:(%d,%d)->(%d,%d)\n", k ++, path[t.x][t.y].x, path[t.x][t.y].y, t.x, t.y);            if(edge[t.x][t.y] != '.') {                int fight = edge[t.x][t.y] - '0';                while(fight --) {                    printf("%ds:FIGHT AT (%d,%d)\n", k ++, t.x, t.y);                }            }        }    }    puts("FINISH");    return;}int main(){    while(~scanf("%d%d", &n, &m)) {        memset(vis, 0, sizeof(vis));        for(int i = 0; i < n; ++ i) {            scanf("%s", edge[i]);        }        Node start; //起点        start.x = start.y = start.step = 0;        minStep = bfs(start);        print(); //打印路径    }    return 0;}